+0

# plz help asap

0
56
3
+18

Find constants A and B such that $$\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}$$
for all x such that$$x\neq -1 and x\neq 2.$$Give your answer as the ordered pair (A,B).

Jul 30, 2020

#1
0

If you let x go infinity, then you get 0 = A + B.

If you let x = 0, then you get -7/2 = A/(-2) + B.

The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3).

Jul 30, 2020
#2
+30926
+1

Hmm, not quite!  if you let x go to infinity then you get 0 = 0.

Let x go to 0 is fine.  Probably better to let x go to 1 rather than infinity for another choice though.

Jul 30, 2020
#3
+18
+2

wasting lots of sticky notes i finally got it

$$\frac{x+7}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}$$ now mulitply (x-2)(x+1) on both sides

$$x+7=Ax+A+Bx-2B$$ trying an example of x, i use 0 which gives

$$7=A-2B$$ and do another that is 4

$$11=5A+2B$$ the I add both equations together

$$18=6A$$ which leaves us with A=3 plugging this into the equation

$$7=3-2B$$ and so B equals -2 leaving us with

the ordered pair for (A,B) as $$3,-2$$

Jul 30, 2020